An aeroplane lands on the runway at velocity 180km/hour and decelerates at 0.4 m/s^2
find the time taken until it stops and the displacement.
Vo = 180000,/3600s. = 50 m/s.
a. V = Vo + a*t = 0.
t =-Vo/a = -50/-0.4 = 125 s.
b. D = Vo*t + 0.5a*t^2.
D = 50*125 - 0.2*2500 = 3125 m.
To find the time taken until the aeroplane stops and the displacement, we can use the equations of motion.
The equation relating velocity, acceleration, and time is:
v = u + at, where
v = final velocity
u = initial velocity
a = acceleration
t = time taken
Given:
Initial velocity (u) = 180 km/hour
Deceleration (a) = -0.4 m/s^2 (negative sign indicates deceleration)
Final velocity (v) = 0 m/s (as the aeroplane stops)
We need to convert the initial velocity from km/hour to m/s.
First, let's convert the initial velocity from km/h to m/s:
1 km/h = (1 * 1000) meters / 3600 seconds = 0.2778 meters / second.
Initial velocity (u) = 180 km/hour * 0.2778 m/s = 50 m/s.
Now that we have the initial velocity (u), deceleration (a), and final velocity (v), we can find the time taken using the equation of motion:
v = u + at.
Plugging in the values:
0 = 50 + (-0.4)t.
Rearranging the equation to solve for t:
-0.4t = -50.
Dividing both sides by -0.4:
t = -50 / -0.4.
t = 125 seconds.
So, the time taken until the aeroplane stops is 125 seconds.
To find the displacement, we use the equation:
s = ut + 0.5at^2, where
s = displacement.
Plugging in the values:
s = 50 * 125 + 0.5 * (-0.4) * (125)^2.
s = 6250 - 3125.
s = 3125 meters.
Therefore, the displacement until the aeroplane stops is 3125 meters.